C# windows form application-PLEASEE HELLLPPP!!!

Question

need help on making a windows form application with a numericupdown and listbox button on it. when I enter 2 in numericupdown button, it should write 1,2 2,1 in listbox. or when I enter 3 in numericupdown, it should create 1,2,3
1,3,2
2,1,3
2,3,1
3,1,2
3,2,1
it is like a binary tree, creating all possible options. please help!!!

Vulpes · Answer

Here's a version which uses a recursive algorithm to generate the permutations. Again I wouldn't use a maximum value of more than 8 for the NumericUpDown:
private List GetPermsRecursive(int n)         {             if (n             List perms = new List ();             int[] a = new int[n];             Visit(a, n, 0, -1, perms); // call to recursive method             return perms;         }
private void Visit(int[] a, int n, int k, int level, List perms)         {             level++;             a[k] = level;
if (level == n)             {                 perms.Add(DisplayPerm(a));             }             else             {                 for (int i = 0; i                 {                     if (a[i] == 0) Visit(a, n, i, level, perms);                                    }             }
level--;               a[k] = 0;          }
private string DisplayPerm(int[] a)         {             var sb = new System.Text.StringBuilder();
for (int i = 0; i             {                 sb.Append(a[i].ToString());                 if (i             }
return sb.ToString();         }
private void numericUpDown1_ValueChanged(object sender, EventArgs e)         {             int n = (int)numericUpDown1.Value;             listBox1.DataSource = GetPermsRecursive(n);         }

musa demirtas · Answer

how can I use recursive method to do the same thing???

Vulpes · Answer

Try the following which uses the QuickPerm algorithm (I've translated it from C to C#) to calculate all permutations of the numbers 1,2, .., n:
http://www.freewebs.com/permute/quickperm.html
I'd keep the maximum value of the NumericUpDown control to a fairly low value (say 8), otherwise it will take a long time to compute all the permutations:
private List QuickPerm(int n)         {             if (n             List perms = new List ();             int[] a = new int[n];             int[] p = new int[n]; // all zero by default              int i, j, tmp;             for (i = 0; i             perms.Add(DisplayPerm(a));             i = 1;             while (i             {                 if (p[i]                 {                     j = i % 2 * p[i];                     tmp = a[j];                     a[j] = a[i];                     a[i] = tmp;                     perms.Add(DisplayPerm(a));                     p[i]++;                     i = 1;                 }                 else                 {                     p[i] = 0;                     i++;                 }             }
return perms;         }
private string DisplayPerm(int[] a)         {             var sb = new System.Text.StringBuilder();
for (int i = 0; i             {                 sb.Append(a[i].ToString());                 if (i             }
return sb.ToString();         }
private void numericUpDown1_ValueChanged(object sender, EventArgs e)         {             int n = (int)numericUpDown1.Value;             listBox1.DataSource = QuickPerm(n);         }